Gravitational potential of a sphere

AI Thread Summary
The discussion focuses on understanding the gravitational potential of a solid sphere versus a hollow sphere. It highlights that inside a solid sphere, gravitational effects are derived from the core, while in a hollow sphere, gravity is effectively zero. The participants discuss the equations for gravitational force, noting that outside the sphere, the force follows the inverse square law (g = GM/x²), while inside, it is proportional to the radius (g = GMx/R³). The conversation draws parallels to Gauss' Law in electrostatics to explain these gravitational principles. Ultimately, the key takeaway is the relationship between mass distribution and gravitational effects in different regions of a sphere.
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Homework Statement


So in this video: https://www.youtube.com/watch?v=rm3x2X0X_Sc&t=210 Why does g.out and g.in have values as shown on the video? I can not for life of my understand it.



Homework Equations





The Attempt at a Solution

 
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Its a solid sphere so as you pass through the surface and approach the center, the gravitational effects are from the core of the sphere not the shell above you.

In a hollow sphere, the effect of gravity is zero.

You can think of a solid sphere as a collection of shells fitting inside each other.
 
jedishrfu said:
Its a solid sphere so as you pass through the surface and approach the center, the gravitational effects are from the core of the sphere not the shell above you.

In a hollow sphere, the effect of gravity is zero.

You can think of a solid sphere as a collection of shells fitting inside each other.

yeah but I still don't understand why it's \frac{GM}{x^{2}} in x>R case and \frac{GMx}{R^{3}} in the other one. I have no idea where the x^2 go and r^3 and x came

Please just help me understand it, I'll never have physics again in my life after tommorows test.
 
Okay nvm I get it, first one was just multiplied by a ratio of what is generating the gravitational effect for our current position, so \frac{x}{R}, am I right?
 
If you consider the sphere as being composed of many smaller spheres of radius x, then each one has volume (4/3)πx3. The total mass of the solid sphere is M. If we suppose a uniform sphere, the mass of a sphere of radius x, M(x), is given by M x3/R3.

Sub this into the expression g(x) = G M(x)/x2. In this way you can derive the field outside too using continuity. If you've ever done electrostatics, this is basically an analogous version of Gauss' Law for gravity.
 
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